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Area of similar triangles

Theorem 6.6
Important
Deleted for CBSE Board 2023 Exams

Ex 6.4, 9 (MCQ) Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 Important Deleted for CBSE Board 2023 Exams

Ex 6.4, 4 Deleted for CBSE Board 2023 Exams

Ex 6.4, 8 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 6.4, 2 Deleted for CBSE Board 2023 Exams

Ex 6.4, 6 Important Deleted for CBSE Board 2023 Exams

Example 9 Important Deleted for CBSE Board 2023 Exams

Ex 6.4, 7 Important Deleted for CBSE Board 2023 Exams

Ex 6.4, 5 Important Deleted for CBSE Board 2023 Exams

Ex 6.4, 3 Important Deleted for CBSE Board 2023 Exams You are here

Last updated at June 23, 2017 by Teachoo

Ex 6.4, 3 In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that (ππ(π΄π΅πΆ))/(ππ(π·π΅πΆ)) = π΄π/π·π Given: β ABC and β DBC Having common base BC To prove: (ππ β π΄π΅πΆ)/(ππ β π·π΅πΆ)=π΄π/π·π Proof: Since AO and OD are not part of β π΄π΅πΆ and β π·π΅πΆ , we cannot directly use the theorem We know that Area of triangle = 1/2Γπ΅ππ πΓπ΄ππ‘ππ‘π’ππ Lets draw altitude AE [ AE β₯π΅πΆ ] Hence, ar β ABC=1/2Γπ΅πΆΓπ΄πΈ Similarly, for Ξ DBC Let us draw altitude DF (DF β₯ BC) Hence, ar β DBC=1/2Γπ΅πΆΓπ·πΉ Now, taking ratio (ππ βπ΄π΅πΆ)/(ππ βπ·π΅πΆ) = (1/2 Γ π΅πΆ Γ π΄πΈ)/(1/2 Γ π΅πΆ Γ π·πΉ) (ππ βπ΄π΅πΆ)/(ππ βπ·π΅πΆ) = π΄πΈ/π·πΉ Now in β AOE and Ξ DOF β AEO = β DFO β AOE = β DOF So, by using AA similarity criterion β AOE ~ β DOF If two triangle are similar their , corresponding sides are in the same ratio So, π΄πΈ/π·πΉ=π΄π/π·π Putting (2) in (1) (ππππ ππ β π΄π΅πΆ)/(ππππ ππ β π·π΅πΆ)=π΄πΈ/π·πΉ (ππππ ππ β π΄π΅πΆ)/(ππππ ππ β π·π΅πΆ)=π΄π/π·π Hence proved